Let `a= lim _(x rarr 1) (x/(lnx)-1/(xln x)), b = lim _(x rarr 0) ((x^(3)-16x)/(4x+x^(2))),`
`c= lim _(x rarr 0) ((ln(1+sinx))/x) & `
`d = lim _(x rarr -1) ((x+1)^(3))/(3([sin (x+1) - (x+1)]))`
Then `[[a,b],[c,d]]` is

Options

A. idempotent
B. involutory
C. non-singular
D. nilpotent

Answer

D

Explanation

`because a lim_(xrarr1) (x/(ln x) - 1/(x ln x)) = lim _(x rarr 1)((x^(2)-1)/(x ln x))`
`= lim_(xrarr 1) ((2x)/(1+lnx))` [by L 'Hospital' s Rule]
`b= lim_(xrarr 0) ((x^(3) - 16 x)/(4x+x^(2)))`
`= lim_(xrarr 0) ((x(x+4)(x-4))/(x(x+4))) = lim_(xrarr 0 ) (x-4) = -4`
`c= lim_(xrarr 0 ) (ln(1+sinx))/x`
`= lim_(xrarr 0 ) (ln(1+sinx))/x cdot lim _(x rarr0) (sin x)/x = 1 cdot 1 = 1`
`= lim _(x rarr -1) ((x+1)^(3))/(3 [ sin (x+1) - (x+1)])`
`= lim _(x rarr -1) (3(x+1)^(3))/(3 [ cos (x+1) - 1])`
[ using L' Hospital' s Rule]
`= lim _(x rarr -1) 1/(([1-cos (x+1)])/((x+1)^(2)) ) = -2`
Let `A= [[2,-4],[1,-2]]rArr A^(2)=0`